Ive checked my code and everything looks fine. Here is the code below. Please tell me why i'm getting this error!Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/adriantr/public_html/joke.php on line 6
Code: Select all
<?php
include("admin/config.php");
$jokeid = $_GET["jokeid"];
$sql = "";
$result = mysql_query("SELECT * FROM news WHERE jokeid=$jokeid");
$row = mysql_fetch_array($result);
$joketitle = $row["joketitle"];
$thejoke = $row["thejoke"];
$jokeimgurl = $row["jokeimgurl"];
$jokeviews = $row["jokeviews"];
$jokeviews = $jokeviews + 1;
echo("<table width=\"500\" border=\"0\" cellspacing=\"00\" cellpadding=\"5\">
<tr>
<td width=\"50\"><img src=$jokeimgurl width=40 height=40></td>
<td>$joketitle - Date Added: $jokedate - Number of Views: $jokeviews </td>
</tr>
<tr>
<td></td>
<td>$thejoke</td>
</tr>
</table>");
mysql_close();
$sql = "UPDATE jokes SET jokeviews='$jokeviews' WHERE jokeid=$jokeid";
$result = mysql_query($sql);
?>-Adrian